Dear Jess Wellendorff
a) I followed your instruction to study the adsorption energy of H2O on calcite.
Firstly, one water molecule is added to the relaxed calcite surface (bottom two layers fixed).
Then, a MD simulation with the NVT ensemble was done from 300k to 1k. (total time 200ps; time step 1ps). I consider the system achieve equilibrium.
The adsorption energy i got is -0.95eV similar to the result (-1 eV )published in Journal of Physical Chemistry C 118(6):3078-3087 (2014).
calcite+H2O= -40556.574 eV
calcite= -40088.730 eV
H2O=-466.867 eV
dE=-0.95eV.
b) However, if i further do the LBFGS geometry optimization for this equilibrium system.
i got calcite+H2O = -40557.62 eV, which means adsorption energy of -~2eV. I know this step is not suggested in the previous discussion.
http://quantumwise.com/forum/index.php?topic=4148.msg19013#msg19013But this way is similar to the tutorial of CO/Pd,
Could you help to advise why i get a different equilibrium state with a lower adsorption energy around -2.0 eV? ( which means more stable ? or something else).
c) In the beginning, i thought this is due to i didn't use proper boundary condition to remove dipole effect. So i use FFT2D with right-sided Dirichlet boundary and left-sided Neumann boundary to calculate the system again.
First, I calculate 4layers calcite again by using LBFGS geometry optimization and fixing the bottom 2 layers. Same as previous case i did, but with a new boundary condition.
When i got surface relaxed calcite, i added one water molecule to the surface of calcite.
Then, i directly use LBFGS geometry optimization to simulate the system and got a similar result of -2.4 eV.
The details of total energy for the system is as follows
calcite+H2O= -40551.210 eV
calcite= -40081.950 eV
H2O=-466.867 eV
dE=-2.393eV.
I attached script of case a and c for your reference.
log file it too big to attach it.
Thanks in advance.